Question

3x^2+x-17 <0 aaaaaaa help

Answer 1

Получается какая- то дичь, может условие неправильно написали:
y = 3x^2+x-17
3x^2+x-17=0
D=1+204=205
x1,2= (-1+-корень из 205)/6
((-1-кор из 205)/6 ; (-1+кор из 205)/6)

Answer 2

To solve the inequality $3x^2 + x - 17 < 0$, we need to find the intervals of $x$ for which the expression is less than zero. Here's how you can do it step by step:

1. Factor the Quadratic Expression: Unfortunately, the quadratic $3x^2 + x - 17$ cannot be easily factored. So, we'll use the quadratic formula to find its roots and then proceed.

2. Quadratic Formula: The roots of the quadratic equation $ax^2 + bx + c = 0$ can be found using the quadratic formula:

   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

   For your quadratic equation $3x^2 + x - 17 = 0$, $a = 3$, $b = 1$, and $c = -17$.

   The discriminant ($b^2 - 4ac$) in this case is $1^2 - 4(3)(-17) = 1 + 204 = 205$, which is positive.

   So, the roots are: $x = \frac{-1 + \sqrt{205}}{6}$ $x = \frac{-1 - \sqrt{205}}{6}$

3. Find the Intervals: To solve the inequality $3x^2 + x - 17 < 0$, we need to find the intervals where the quadratic expression is less than zero.

   Since the coefficient of $x^2$ is positive (3), the quadratic opens upwards, forming a U-shape. The expression is negative between the two roots, so we're interested in the interval where $x$ lies between these roots.

   The values of the roots are approximately: $x_1 \approx 2.253$ $x_2 \approx -2.986$

4. Inequality Solution: Now we need to find the interval for $x$ where the expression $3x^2 + x - 17$ is negative. This interval is $(-2.986, 2.253)$.

Therefore, the solution to the inequality $3x^2 + x - 17 < 0$ is: $x \in (-2.986, 2.253)$

This means that the inequality is satisfied when $x$ is within the interval $(-2.986, 2.253)$.